How do u solve this?

Answer:

A. 5

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Algebra II

Step-by-step explanation:

Step 1: Define

Identify points from graph.

Point (8, 5)

Point (4, 2)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Answer:

I think the answer is B. 65%

Answer:

[tex]Pr= \frac{1}{21}[/tex]

Step-by-step explanation:

Given

[tex]n(S) = 7[/tex] --- number of games

Required

Probability of bike and movie in consecutive order

This probability is represented as:

[tex]Pr = P(Bike\ and\ Movie) \ or\ P(Movie\ or\ Bike)[/tex]

So, we have:

[tex]Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)[/tex]

The probability is an illustration of selection without replacement;

So, we have:

[tex]P(Bike\ and\ Movie) = P(Bike) * P(Movie)[/tex]

[tex]P(Bike\ and\ Movie) = \frac{n(Bike)}{n(S)} * \frac{n(Movie)}{n(S) - 1}[/tex] ---- without replacement

Bike and Movie appear in the game list 1 time.

So, the equation becomes

[tex]P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{7 - 1}[/tex]

[tex]P(Bike\ and\ Movie) = \frac{1}{7} * \frac{1}{6}[/tex]

[tex]P(Bike\ and\ Movie) = \frac{1}{42}[/tex]

Similarly,

[tex]P(Movie\ and\ Bike) = \frac{1}{42}[/tex]

So, we have:

[tex]Pr = P(Bike\ and\ Movie) \ +\ P(Movie\ or\ Bike)[/tex]

[tex]Pr= \frac{1}{42}+\frac{1}{42}[/tex]

Take LCM

[tex]Pr= \frac{1+1}{42}[/tex]

[tex]Pr= \frac{2}{42}[/tex]

[tex]Pr= \frac{1}{21}[/tex]

Answer:

c

Step-by-step explanation:

The first four terms are 17, 15, 13, and 11.

A function is a relationship between inputs where each input is related to exactly one output.

Example:

f(x) = 2x + 1

f(1) = 2 + 1 = 3

f(2) = 2 x 2 + 1 = 4 + 1 = 5

The outputs of the functions are 3 and 5

The inputs of the function are 1 and 2.

We have,

f(n):

f(1) = 17

f(n) = f(n - 1) - 2 if n > 1

Now,

The first term is 17.

The second term.
f(2) = f(2 - 1) - 2

= f(1) - 2

= 17 - 2

= 15

The third term.

= f(3 - 1) - 2

= f(2) - 2

= 15 - 2

= 13

The fourth term.

f(4) = f(4 - 1) - 2
= f(3) - 2

= 13 - 2

= 11

Thus,

The terms are 17, 15, 13, 11

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Answer:

option D is the answer

Step-by-step explanation:

using the HH,ll,ha and la,

where h is the hypotenuse and l is the leg

Answer:

4^15

Step-by-step explanation:

We know a^b^c = a^(b*c)

4^3^5

4^(3*5)

4^15

Answer:

12

Step-by-step explanation:

If f(x) = 5 - x

Then f(-7) = 5 - (-7)

f(-7) = 5 + 7

f(-7) = 12

Answer:

Horizontal Shift: Right 1

Vertical Shift: Down 5

Reflection: None

Explanation: To find the transformation, compare the function to the parent function (being in this case g(x)=1/x) and check to see if there is a horizontal or vertical shift or a reflection.

So, the answer would be Right 1, and down 5

Hope this helps you out :)

Answer:

a) 0.2

b) 0.0009

c) 0.0298

d) 0.0465 = 4.65% probability that the sample proportion is less than 0.15.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

20% of customers entering her store make a purchase.

This means that [tex]p = 0.2[/tex]

180 people

This means that [tex]n = 180[/tex]

a. What is the mean of the distribution of the sample proportion of customers making a purchase?

By the Central Limit Theorem, [tex]\mu = p = 0.2[/tex].

b) What is the variance of the sample proportion?

The standard deviation is:

[tex]s = \sqrt{\frac{0.2*0.8}{180}} = 0.0298[/tex]

Variance is the square of the standard deviation, so:

[tex]s^2 = (0.0298)^2 = 0.0009[/tex]

c) What is the standard error of the sample proportion?

As found in the previous item, 0.0298.

d) What is the probability that the sample proportion is less than 0.15?

This is the p-value of Z when X = 0.15. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.15 - 0.20}{0.0298}[/tex]

[tex]Z = -1.68[/tex]

[tex]Z = -1.68[/tex] has a p-value of 0.0465.

0.0465 = 4.65% probability that the sample proportion is less than 0.15.

(1) D

[tex]L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt[/tex]

Integrate by parts, taking

[tex]u = t \implies \mathrm du=\mathrm dt[/tex]

[tex]\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}[/tex]

Then

[tex]L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}[/tex]

[tex]L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}[/tex]

(2) A

[tex]L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt[/tex]

[tex]L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}[/tex]

[tex]L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}[/tex]

Answer:

A.  5

Step-by-step explanation:

Parallel lines have the same slope.

Answer:

5

Step-by-step explanation:

Answer:

a. 168000 for Eyeconnect, 132000 for Topconnect

b. 100,000

Step-by-step explanation:

a.

Because the change in customers are only due to leaving companies, we can say that, after one year, Eyeconnect loses 10% of its customers to Topconnect and Topconnect loses 5% of its customers to Eyeconnect. This represents all changes in customers.

First, we can calculate how much Eyeconnect loses, which is 10% of 180,000 = 0.1 * 180,000 = 18,000 . They then have 180,000 - 18,000 = 162,000 employees

Next, Topconnect loses 120,000 * 5% = 120,000 * 0.05 = 6,000. They then have 120,000-6,000 = 114,000 employees

We can then add the customer amounts. Note that we are subtracting both sides before adding as both companies gain and lose customers simultaneously.

We can then add how much one company lost to the other company's customers.

Eyeconnect gains 6,000 customers, so they then have 162,000 + 6,000 = 168000 employees. Topconnect gains 18,000 customers so they then have 114,000 + 18,000 = 132,000 employees

b.

After many years, the number of customers Eyeconnect has will be less than the number of customers that Topconnect has. One way to find the end amount of customers that Eyeconnect has is to figure out when the customer bases even out, or when Eyeconnect loses the same amount of customers as Topconnect so the customer base stays the exact same. We know that no customers leave or join the companies except to leave/join the other, so the total amount of customers between the two companies stays the exact same. The amount of customers is 180,000 + 120,000 = 300,000. Therefore, at the end amount,

Eyeconnect customers (E) + Topconnect customers (T) =300,000

Furthermore, if the amount of customers that leave Eyeconnect is the same that leaves Topconnect, we can say

E * 0.1 = T * 0.05

divide both sides by 0.05 to isolate the T

E * 0.1 / 0.05 = T

2 * E = T

plug that into the first equation

E + 2 * E = 300,000

3 * E = 300,000

divide both sides by 3 to isolate E

E = 100,000 after many years

Answer:

the write answer to your question is tan 25 degree

With some rewriting, you get

[tex]\displaystyle \sum_{k=0}^\infty (-1)^k\frac{x^{k+2}}{4^k} = x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k[/tex]

Recall that for |x| < 1, you have

[tex]\displaystyle \frac1{1-x} = \sum_{k=0}^\infty x^k[/tex]

So as long as |-x/4| = |x/4| < 1, or |x| < 4, your series converges to

[tex]\displaystyle x^2 \sum_{k=0}^\infty \left(-\frac x4\right)^k = \frac{x^2}{1-\left(-\frac x4\right)} = \frac{x^2}{1+\frac x4} = \boxed{\frac{4x^2}{4+x}}[/tex]

Based on known expressions from Taylor series, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex].

Taylor series are polynomic approximations used to estimate values both from trascendental and non-trascendental functions. It is commonly used in trigonometric, potential, logarithmic and even rational functions.

In this question we must use series properties and common Taylor series-derived formulas to infer the expression behind the given series. Now we proceed to find the expression:

[tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex]

[tex]x^{2}\cdot \sum\limits_{k = 0}^{\infty} \left(-\frac{x}{4} \right)^{k}[/tex]

[tex]x^{2}\cdot \left(\frac{1}{1+\frac{x}{4} } \right)[/tex]

[tex]\frac{4\cdot x^{2}}{4+x}[/tex]

Based on power and series properties and most common Taylor series- derived formulas, the power series [tex]\sum \limits_{k = 0}^{\infty} (-1)^{k}\cdot \frac{x^{k+2}}{4^{k}}[/tex] represents a Taylor series-derived formula of the rational function [tex]\frac{4\cdot x^{2}}{4+x}[/tex]. [tex]\blacksquare[/tex]

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Answer:

D. 28

Step-by-step explanation:

Given:

m∠AFD = 90°

m∠AFB = 31°

Required:

m∠DFE

Solution:

m<AFB = m<CFD (both angles are marked as congruent angles)

Since m<AFB = 31°, therefore,

m<CFD = 31°

m<AFB + m<CFD + m<BFC = m<AFD

Plug in the known values

31° + 31° + m<BFC = 90°

62° + m<BFC = 90°

Subtract 62° from each side

m<BFC = 90° - 62°

m<BFC = 28°

m<BFC = m<DFE = 28° (both angles are marked congruent to each other)

Therefore,

m<DFE = 28°

Answer:

(a)The probability that both marbles are blue=5/18

The probability that both marbles are yellow=1/6

(b)The probability that exactly one marble is blue​=5/9

Step-by-step explanation:

Blue marbles=5

Yellow marbles=4

Total marbles=5+4=9

(a)

Probability of drawing first  blue marble=5/9

Probability of drawing second blue marble without replacement=4/8

The probability that both marbles are blue

[tex]=\frac{5}{9}\times \frac{4}{8}=\frac{5}{18}[/tex]

Probability of drawing first  yellow marble=4/9

Probability of drawing second yellow marble without replacement=3/8

The probability that both marbles are yellow

[tex]=\frac{4}{9}\times \frac{3}{8}=\frac{1}{6}[/tex]

(b)

The probability that exactly one marble is blue​

=Probability of first blue marble (Probability of second yellow marble)+Probability of first yellow marble (Probability of second blue marble)

The probability that exactly one marble is blue​

=[tex]\frac{5}{9}\times \frac{4}{8}+\frac{4}{9}\times \frac{5}{8}[/tex]

=[tex]\frac{5}{18}+\frac{5}{18}[/tex]

=[tex]\frac{10}{18}=\frac{5}{9}[/tex]

Answer:

x = 38.4

Step-by-step explanation:

tan(38) = 30/x

x = 30/tan(38)

x = 38.4

Answered by GAUTHMATH

========================================================

Explanation:

You can use the AAS (angle angle side) theorem to prove that triangle ABD is congruent to triangle CBD.

From there, we can then say that AD and DC are the same length

AD = DC

3y+6 = 5y-18

3y-5y = -18-6

-2y = -24

y = (-24)/(-2)

y = 12

Answer:

i believe it'll cost 200 dollars

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

[tex]\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42[/tex]

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

So

[tex]s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772[/tex]

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is [tex]n = 10[/tex]

The significance level is [tex]\alpha = 1 - 0.05 = 0.95[/tex]

The estimate, which is the sample standard deviation, is of [tex]s = 0.8772[/tex].

Now, we have to find the critical values for the Pearson distribution. They are:

[tex]\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228[/tex]

[tex]\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004[/tex]

The confidence interval for the population variance is:

[tex]\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}[/tex]

[tex]\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}[/tex]

[tex]0.3641 < \sigma^2 < 2.5646[/tex]

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

[tex]\sqrt{0.3641} = 0.6034[/tex]

[tex]\sqrt{2.5646} = 1.6014[/tex]

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

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Step-by-step explanation:

X= - 1

X=0

X=4/3

X=3

We solve for x by simplifying both sides of the equation, then isolate the variable.

Answer :

C (x=4/3)

9514 1404 393

Answer:

  13 < √181 < 14

Step-by-step explanation:

Apparently, you're supposed to know that ...

  13² = 169

  14² = 196

so √181 will lie between 13 and 14.

  13 < √181 < 14

Answer:

2

Step-by-step explanation:

8 ^ (5/3) ^ 1/5

We know a^b^c = a^(b*c)

8^ (5/3*1/5)

8^ 1/3

Rewriting 8 as 2^3

2^3 ^1/3

2 ^(3*1/3)

2^1

2

Answer:

2

Step-by-step explanation:

((2^3)^5/3)^1/5

= (2^5)^1/5

= 2

Answered by Gauthmath

[tex]\boxed{ \sf{Answer}} [/tex]

[tex] \sf \: 3(3x - 4) - 5(x + 3) \\ \sf =( 3 \times 3x) - (3 \times 4) + ( - 5 \times x) +( - 5 \times 3 ) \\ \sf = 9x - 12 - 5x - 15 \\ \sf = 9x - 5x - 12 - 15 \\ = \underline{ \bf 4x - 27}[/tex]

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐

[tex]\\ \sf\longmapsto 3(3x-4)-5(x+3)[/tex]

[tex]\\ \sf\longmapsto 9x-12-5x-15[/tex]

[tex]\\ \sf\longmapsto 9x-5x-15-12[/tex]

[tex]\\ \sf\longmapsto (9-5)x-27[/tex]

[tex]\\ \sf\longmapsto 4x-27[/tex]

Answer:

$199.29

Step-by-step explanation:

Total payments = $7,174.24

Total interest = $1,174.24

Answer:

12024.7

Step-by-step explanation:

Searched it up.

then rounded

The required probability of a couple having a baby boy when their third child is​ born is 1/2.

probability is the ratio of the number of favorable outcomes and the total number of possible outcomes. The chance that a particular event (or set of events) will occur expressed on a linear scale from 0 (impossibility) to 1 (certainty), also expressed as a percentage between 0 and 100%.

Given:

Assuming boys and girls are equally​ likely.

The first two children were both boys

According to given question we have

The probability of having a baby girl is an independent probability.

The first two children were both boys

So, it is not related to the previous child.

So required probability = 1/2

Therefore, the required probability of a couple having a baby boy when their third child is​ born is 1/2.

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It looks like the integral in polar coordinates is given to be

[tex]\displaystyle\int_{\pi/2}^\pi \int_0^2 r^3\sin(\theta)\cos(\theta)\,\mathrm dr\,\mathrm d\theta[/tex]

Converting back to Cartesian, we take

x = r cos(θ)

y = r sin(θ)

dx dy = r dr dθ

so we can easily recover the integrand in Cartesian:

[tex]r^3\sin(\theta)\cos(\theta)\,\mathrm dr\,\mathrm d\theta = (r\sin(\theta))(r\cos(\theta))(r\,\mathrm dr\,\mathrm d\theta) = xy\,\mathrm dx\,\mathrm dy[/tex]

This leaves us with the limits:

• π/2 ≤ θ ≤ π corresponds to the second quadrant of the (x, y)-plane (that is, where x < 0 and y > 0)

• 0 ≤ r ≤ 2 correspond to the disk of radius 2 centered at the origin

Taken together, we see the region of integration is a quarter-disk of radius 2 in the second quadrant, which we can capture by the set

{(x, y) : -√2 ≤ x ≤ 0 and 0 ≤ y ≤ √(2 - x ²)}

So, in Cartesian coordinates, the integral would be

[tex]\displaystyle \boxed{\int_{-\sqrt2}^0 \int_0^{\sqrt{2-x^2}} xy\,\mathrm dy\,\mathrm dx}[/tex]

Answer:

x = 8/3

Step-by-step explanation:

Log₂(9x) – Log₂3 = 3

The value of x can be obtained as follow:

Log₂(9x) – Log₂3 = 3

Recall

Log M – Log N = Log (M/N)

Thus,

Log₂(9x) – Log₂3 = 3

Log₂(9x/3) = 3

Log₂3x = 3

3x = 2³

3x = 8

Divide both side by 3

x = 8/3

Answer:

the first option because I took the test